Problem: You have found the following ages (in years) of all 4 snakes at your local zoo: $ 9,\enspace 8,\enspace 21,\enspace 1$ What is the average age of the snakes at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we have data for all 4 snakes at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{9 + 8 + 21 + 1}{{4}} = {9.8\text{ years old}} $ Find the squared deviations from the mean for each snake. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $9$ years $-0.8$ years $0.64$ years $^2$ $8$ years $-1.8$ years $3.24$ years $^2$ $21$ years $11.2$ years $125.44$ years $^2$ $1$ year $-8.8$ years $77.44$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{0.64} + {3.24} + {125.44} + {77.44}} {{4}} $ $ {\sigma^2} = \dfrac{{206.76}}{{4}} = {51.69\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{51.69\text{ years}^2}} = {7.2\text{ years}} $ The average snake at the zoo is 9.8 years old. There is a standard deviation of 7.2 years.